If you are playing tug of war with your friend and you move your friend 3 meters using a force of 5 N, How much work did you do?

If a 28.0 L balloon with a temperature of

Alexeev081 [22]

Answer:

5.6 L

Explanation:

We can apply Charles' Law here since our pressure is constant (will not change inside the refrigerator) and we are relating change in volume with change in temperature:

V₁ / T₁ = V₂ / T₂ where V₁ and T₁ are initial volume and temperature, and V₂ and T₂ are final volume and temperature. Let's plug in what we know and solve for the unknown:

28.0 L / 25 °C = V₂ / 5 °C => V₂ = 5.6 L

5.6 L is our new volume (at 5 °C).

6 0

2 years ago

What is the mass of 1.5 moles of CO2?

Arisa [49]

1 mole of CO2 has 44g
so 1,5 moles would have 1,5*44=66g
the answer is 1) 66g

8 0

3 years ago

For question #12, use the following picture:<br><br> 12. What element is this? How do you know?

labwork [276]

Answer: The element shown in the image is Helium (He).

Explanation: We are given a image of an atom having protons, neutrons and electrons.

Number of protons as shown in image = 2

Number of neutrons as shown in image = 2

Number of electron as shown in image = 2

Atomic number = Number of protons = Number of electrons

Atomic number of the element = 2

Atomic Mass = Number of protons + Number of neutrons

Atomic mass = 2 + 2 = 4

The element having Atomic number = 2 and mass number = 4 is Helium.

Element = $^4_2\textrm{He}$

8 0

3 years ago

Read 2 more answers

In methane combustion, the following reaction pair is important: At 1500 K, the equilibrium constant Kp has a value of 0.003691

andre [41]

Explanation:

Let us assume that the value of $K_{r}$ = $2.82 \times 10^{5} \times e^({\frac{-9835}{T}}) m^{6}/mol^{2}s$

Also at 1500 K, $K_{r}$ = $2.82 \times 10^{5} \times e^({\frac{-9835}{1500}}) m^{6}/mol^{2}s$

$K_{r} = 400.613 m^{6}/mol^{2}s$

Relation between $K_{p}$ and $K_{c}$ is as follows.

$K_{p} = K_{c}RT$

Putting the given values into the above formula as follows.

$K_{p} = K_{c}RT$

$0.003691 = K_{c} \times 8.314 \times 1500$

$K_{c} = 2.9 \times 10^{-7}$

Also, $K_{c} = \frac{K_{f}}{K_{r}}$

or, $K_{f} = K_{c} \times K_{r}$

= $2.9 \times 10^{-7} \times 400.613$

= $1.1617 \times 10^{-4} m^{6}/mol^{2}s$

Thus, we can conclude that the value of $K_{f}$ is $1.1617 \times 10^{-4} m^{6}/mol^{2}s$ .

6 0

2 years ago

Salt is often added to water to raise the boiling point to heat food more quickly. if you add 30.0g of salt to 3.75kg of water,

sammy [17]

Assuming an ebullioscopic constant of 0.512 °C/m for the water, If you add 30.0g of salt to 3.75kg of water, the boiling-point elevation will be 0.140 °C and the boiling-point of the solution will be 100.14 °C.

<h3>What is the boiling-point elevation?</h3>

Boiling-point elevation describes the phenomenon that the boiling point of a liquid will be higher when another compound is added, meaning that a solution has a higher boiling point than a pure solvent.

Step 1: Calculate the molality of the solution.

We will use the definition of molality.

b = mass solute / molar mass solute × kg solvent

b = 30.0 g / (58.44 g/mol) × 3.75 kg = 0.137 m

Step 2: Calculate the boiling-point elevation.

We will use the following expression.

ΔT = Kb × m × i

ΔT = 0.512 °C/m × 0.137 m × 2 = 0.140 °C

where

ΔT is the boiling-point elevation
Kb is the ebullioscopic constant.
b is the molality.
i is the Van't Hoff factor (i = 2 for NaCl).

The normal boiling-point for water is 100 °C. The boiling-point of the solution will be:

100 °C + 0.140 °C = 100.14 °C

Assuming an ebullioscopic constant of 0.512 °C/m for the water, If you add 30.0g of salt to 3.75kg of water, the boiling-point elevation will be 0.140 °C and the boiling-point of the solution will be 100.14 °C.

Learn more about boiling-point elevation here: brainly.com/question/4206205

7 0

2 years ago